probability density of a nonhomogeneous poisson process

Question

I am beginning to study Poisson processes and have come across this question involving a function $\lambda(u) = u + 1$ so that it is a non-homogeneous Poisson process. Let $\tau$ be the time between each arrival. How would I calculate the probability density function of $\tau_1$? I feel like I am not understanding the definitions correctly. What exactly is $\lambda(u)$, and what is its relation to the pdf of $\tau_1$? Also, I am wondering if each $\tau_i$ are independent of each other or not? My thoughts for this second question is that $\tau_i$ are not independent. I would somehow show this with conditional probability: $P(\tau_2 \leq b\ |\ \tau_1 = a)$, but I am unsure how to proceed with the proof.

Answer 1

Answering your questions, in the first place the times are not independent unless the function $ \lambda $ is constant, that is, it is a homogeneous Poisson process. In the case of a non-homogeneous process, the value $ \lambda(u) $ represents the rate at time $ u $.

To find the density of $ T_{1} $, you just have to see that: $$1-P(T_{1}\leqslant t)=P(T_{1}>t)=P(N(t)=0)=e^{-\mu(t)}$$ where: $$\mu(t)=\int_{0}^{t}\lambda(u)\,du$$

Deriving this expression we obtain that:

$$f_{T_{1}}(t)=-\frac{\text{d}}{\text{d}t}P(T_{1}>t)=\lambda(t)e^{-\mu(t)}$$

and this is precisely the density of $ T_{1} $. This is exponential in the case where $ \lambda(u) = \text{const}$. If you wanted to find the joint distribution for example between $ T_ {1} $ and $ T_ {2} $, just find the conditional, that is $P(T_{2}>t_{2}|T_{1}=t_{1})$. Now, since we are stopped at moments $ t_ {1} $, the set where $ T_ {2}> t_ {2} $ is equivalent to the number of events occurring at time $ t_ {1} $ is equal to the number of events at time $ t_ {1} + t_ {2} $, that is: $$P(T_{2}>t_{2}|T_{1}=t_{1})=P(N(t_{1}+t_{2})-N(t_{1})=0|T_{1}=t_{1})=P(N(t_{1}+t_{2})-N(t_{1})=0|N(t_{1})\geqslant 1)=P(N(t_{1}+t_{2})-N(t_{1})=0)=e^{-[\mu(t_{1}+t_{2})-\mu(t_{1})]}$$

Deriving with respect to $ t_ {2} $ we can see that: $$f_{T_{2}|T_{1}}(t_{2}|t_{1})=\lambda(t_{1}+t_{2})e^{-[\mu(t_{1}+t_{2})-\mu(t_{1})]}$$

So you already have an expression for the joint density, which is: $$f_{T_{1},T_{2}}(t_{1},t_{2})=\lambda(t_{1})\lambda(t_{1}+t_{2})e^{-\mu(t_{1}+t_{2})}$$

So they are not independent. Notice that we have used throughout the proof that $ N (t) $ has independent increments, and that $ N(t+s)-N(t) $ has a Poisson distribution with parameter $\mu(t+s)-\mu(t)$